兩條中4a.math（好難）（今晚要）

2007-11-22 8:08 am

1. if cosX and cosecX are the two roots of the equation 3x^2+hx-1=0
where 90*<180*
find the values of h (leave ans in surd form)

2.if sinX and cosX are the two roots of equation kx^2-4x+3=0
find the possible values of k

寫明點做點解
急要！
邊個第一個答到有２０分！
我計左唔知幾耐了
都係計唔出

更新1:

1. if cosX and cosecX are the two roots of the equation 3x^2+hx-1=0 where 90*

更新2:

1. if cosX and cosecX are the two roots of the equation 3x^2+hx-1=0 where 180度大過X度大過９０度 find the values of h (leave ans in surd form)

更新3:

錯!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! 比埋兩條ans 第一係 -7/開方根10 第二係 -8

更新4:

你要留意X係１８０度之內　９０度之外　　

更新5:

多謝你提左我有個PRODUCT OF ROOTS 我而家識計啦我地個個呀SIR未教PRODUCT & SUM OF ROOTS 所以我唔識好彩你提左我如果唔係我真係訓唔到教唔該晒！個X係１８０度－７１．５７度

回答 (3)

Kam Kee

2007-11-22 8:50 am

✔ 最佳答案

3x^2+hx-1=0
x^2+(h/3)x+(-1/3)=0

cosX * cosecX = -1/3 (product of roots)
cot X = -1/3
=> tan X = -3
=> cos X = -1/(√10) (自己在第二象限畫個三角形就會明)
=> cosecX = (√10)/3

cosX + cosecX = (-h/3) (sum of roots)
-1/(√10) + (√10)/3 = -h/3
h = -3/(√10) + (√10)

kx^2-4x+3=0
因為 sinX and cosX 係 roots

k(sinX) ^2-4sinX+3=0 --- (1)
k(cosX) ^2-4cosX+3=0 --- (2)

(1) + (2)
k -4sinX+(-4cosX)+ 6 = 0
k = 4(sinX+cosX) - 6

因為 -1 < sinX+cosX < 1
所以 -10 < k < -2

2007-11-22 01:23:09 補充：
第一條通分母再加就係你個答案，恕我漏了 step。第二條我再諗諗，但係我想確定下 possible values of k 多數暗示有好多答案，係唔係 suggest answer 都只係 -8 ?

2007-11-22 01:24:56 補充：
嗯... 明白了，謝謝 ctfa23

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Tom

2007-11-24 6:03 am

3x^2+hx-1=0
x^2+(h/3)x+(-1/3)=0

cosX * cosecX = -1/3 (product of roots)
cot X = -1/3
=> tan X = -3
=> cos X = -1/(√10)
=> cosecX = (√10)/3

cosX + cosecX = (-h/3) (sum of roots)
-1/(√10) + (√10)/3 = -h/3
h = -3/(√10) + (√10)

kx^2-4x+3=0
因為 sinX and cosX 係 roots

k(sinX) ^2-4sinX+3=0 --- (1)
k(cosX) ^2-4cosX+3=0 --- (2)

(1) + (2)
k -4sinX+(-4cosX)+ 6 = 0
k = 4(sinX+cosX) - 6

因為 -1 < sinX+cosX < 1
所以 -10 < k < -2

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TszFung

2007-11-22 9:21 am

1.
樓上的其實只差最後一步,
From product of root
tanX = -3
cosX = -1/root10
cosecX = root10 /3

-h/3 = (-1/root10) + (root10 /3)
h = (3/root10) - root10 = -7 / root10 = -7 root10 /10

2.
sum of roots = sinX + cosX = 4/k
product of roots = sinXcosX = 3/k

both sides take square for sum of roots:
(sinX + cosX)^2 = (4/k)^2
(sinX)^2 + (cosX)^2 + 2sinXcosX = 16 /(k^2)
1 + 2 (3/k) = 16 /(k^2)
k^2 + 6k - 16 = 0
(k+8) (k-2) = 0
k = -8 or 2 (rej.)

k cannot be 2 because sinXcosX (3/k) and sinX + cosX (4/k) cannot be greater than 1

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